For all who liked Ms Misawa's inequalities killer puzzles last year, here is one made by Pyrrhon:

Edited: new pics attached below...

It should be harder than Misawa's ones... Enjoy!

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• Page **1** of **1**

For all who liked Ms Misawa's inequalities killer puzzles last year, here is one made by Pyrrhon:

Edited: new pics attached below...

It should be harder than Misawa's ones... Enjoy!

Edited: new pics attached below...

It should be harder than Misawa's ones... Enjoy!

Last edited by udosuk on Mon Sep 04, 2006 1:24 am, edited 1 time in total.

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

I've verified the 1st version has a unique solution and could be solved with logic (using colors/forcing chains type of moves)...

Also I believe a 2nd version, with a cage removed, should also gives a unique solution, though might not be solvable by a human using logic (what we call RUUDICULOUS puzzles in the killer community)...

I've removed some unnecessary cages and signs, with the following new pics:

Also I believe a 2nd version, with a cage removed, should also gives a unique solution, though might not be solvable by a human using logic (what we call RUUDICULOUS puzzles in the killer community)...

I've removed some unnecessary cages and signs, with the following new pics:

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

A computer program isn't necessary. But there is a nice step which you have eventually overseen. It could be an standard killer technique but I don't know whether JSudoku takes it. I give it as usual in the killer community in tiny text.

Attempt to estimate with help of your information about N5 and 45 rule the lowest possible value of R5C4.Then the puzzle breaks down.

Other puzzles of this variant will eventually come later. In the meantime you can take one of my sum search puzzles at my webpage. This are puzzle of a variant that I invented and that is in it's flavour similar to inequality killer puzzles.

Pyrrhon

Attempt to estimate with help of your information about N5 and 45 rule the lowest possible value of R5C4.Then the puzzle breaks down.

Other puzzles of this variant will eventually come later. In the meantime you can take one of my sum search puzzles at my webpage. This are puzzle of a variant that I invented and that is in it's flavour similar to inequality killer puzzles.

Pyrrhon

- Pyrrhon
**Posts:**240**Joined:**26 April 2006

Here is a walkthrough for puzzle 1 (original version):

(copy and paste the tiny text to Notepad etc to read)

1. In n6, there are 4 cages with the same sum (s1) and the remaining cell has a smaller value

-> only 2 possibilities: 4*11+1 or 4*10+5

Since r67c7<s1, r67c7<=10

But r67c7=r78c89>=10 (sum of 4-cell cage)

-> r67c7=10=[19], r78c89={1234}, s1=11 is the sum of many cages in the grid

-> All 11/2 cages cannot have 1, the 11/3-cage in n2 cannot have 9

-> 1 of n4 locked in r4c12

2. r12c1 < r3c12 = r456c2 < r7c23 = r6c34 = r8c34 < r9c456 < r8c67 < r78c89 = 10

-> r456c2=6=[1{23}] (naked pair), r3c12=6=[15|24]

-> r12c1<6, cannot be {12} or {14} (conflict with r3c12), must be {(12)3}

-> naked triple {123} in r123c1, hidden triple {123} in r789c3

-> r9c456=8={1(25|34)} with 1 locked in, r8c67=9=[27|36|45] (-> naked quad {1234} in r8c3689)

-> r6c34=7=[43|52] (-> naked pair {23} in r6c24), r7c23=7=[43|52|61], r8c34=7=[16|25] (-> HS r8c5=9)

We have an XY-wing here: r8c4=5 will force r8c3=2 and r9c456={134}, leaving nothing for r9c3={23}

Therefore r8c34=[16], r8c67=[27|45], r7c23=[43|52] (-> naked pair {45} in r37c2, HS r7c1=6)

3. r4c34=11=[47|65|74|83|92], r56c6=10=[19|28|37|46|64]

45 on r4 -> r4c17=11 -> r4c1=r5c7=(4578), r4c7=(3467)

45 on n5 -> r456c4+r56c5=24, r56c5<=9 & r456c4>=15 -> r5c4=(5789), r5c5=(12345)

Now r5c5=4|5 would force r56c5=9={45}, but then r456c4 would become {(23789)} unable to give a total of 15

(subtraction combo cannot extract a 14/2 from it) -> r5c5=(123)

Now r4c4 cannot be 2|3, otherwise we'd have a naked triple {123} in r46c4+r5c5, requiring r5c4+r6c5=18, impossible!

-> r4c34=[47|65|74]

Next, we see that there is a strong link of 5 in r8c17, so r4c1=r5c7 cannot both be 5

-> r4c1=r5c7=(478), r4c7=(347)

4. Take a look at r56c6=10:

Case 1: r56c6=[19|28|37] -> naked triple {123} in n5 -> r4c56=11={47|56} -> naked quad of {4567} in r4c3456 -> r45c7=[38]

Case 2: r56c6={46} -> r8c67=[27] -> r45c7=[38]

So no matter what r45c7=[38]!

-> r45c7=[38], r4c1=8, HS r8c2=8

5. Notice that on c1, one of r569c1 must be 4. So r6c3 and r7c2 cannot both be 4, one of them must be 5.

Case 1: r6c3=5 -> r6c4=2 -> r4c56 cannot be {29}

Case 2: r7c2=5 -> r8c1=7 -> r8c67=[45] -> r56c6 cannot be {46} -> r5c6=(123) -> naked triple in n5 -> r4c56 cannot be {29}

So no matter what r4c56 cannot be {29}!

-> r4c56={47|56}, naked quad on r4c3456, r4c89={29}, r56c6=[19|28|37] (hidden triple {123} in n5), 5 in n5 locked in r4c456

6. Since r456c4>=15, if r5c4=7 then r456c4=[573], but that would conflict with r4c56={47|56}

-> r5c4=9 -> HS r6c1=r9c2=9 -> r56c6=[28|37] -> HS r5c5=1 -> n1 contains a naked pair {67} in r12c2

7. {24} on c7 is locked in r123c7 -> r13c6+r2c8 must contain {79} (the 3 11/2 cages) -> r2c456 cannot have {79}

Also, r1c4=7 would force r2c2=7 and rule out any 7 in r13c6+r2c8 -> r1c4<>7

Hence r1c34=11=[83|92] (-> naked pair {23} on r16c4)

One of r13c6 must be 9 -> one of r13c7 must be 2 -> r2c7<>2

8. From step 5 above we know that either r6c3 or r7c2 must be 5:

Case 1: r6c3=5 -> r6c4=2 -> r1c4<>2

Case 2: r7c2=5 -> r3c12=[24] -> r3c7<>2 -> r1c7=2 -> r1c4<>2

So no matter what r1c4<>2!

-> r1c34=[83]

And the rest is all trivial!

(copy and paste the tiny text to Notepad etc to read)

1. In n6, there are 4 cages with the same sum (s1) and the remaining cell has a smaller value

-> only 2 possibilities: 4*11+1 or 4*10+5

Since r67c7<s1, r67c7<=10

But r67c7=r78c89>=10 (sum of 4-cell cage)

-> r67c7=10=[19], r78c89={1234}, s1=11 is the sum of many cages in the grid

-> All 11/2 cages cannot have 1, the 11/3-cage in n2 cannot have 9

-> 1 of n4 locked in r4c12

2. r12c1 < r3c12 = r456c2 < r7c23 = r6c34 = r8c34 < r9c456 < r8c67 < r78c89 = 10

-> r456c2=6=[1{23}] (naked pair), r3c12=6=[15|24]

-> r12c1<6, cannot be {12} or {14} (conflict with r3c12), must be {(12)3}

-> naked triple {123} in r123c1, hidden triple {123} in r789c3

-> r9c456=8={1(25|34)} with 1 locked in, r8c67=9=[27|36|45] (-> naked quad {1234} in r8c3689)

-> r6c34=7=[43|52] (-> naked pair {23} in r6c24), r7c23=7=[43|52|61], r8c34=7=[16|25] (-> HS r8c5=9)

We have an XY-wing here: r8c4=5 will force r8c3=2 and r9c456={134}, leaving nothing for r9c3={23}

Therefore r8c34=[16], r8c67=[27|45], r7c23=[43|52] (-> naked pair {45} in r37c2, HS r7c1=6)

3. r4c34=11=[47|65|74|83|92], r56c6=10=[19|28|37|46|64]

45 on r4 -> r4c17=11 -> r4c1=r5c7=(4578), r4c7=(3467)

45 on n5 -> r456c4+r56c5=24, r56c5<=9 & r456c4>=15 -> r5c4=(5789), r5c5=(12345)

Now r5c5=4|5 would force r56c5=9={45}, but then r456c4 would become {(23789)} unable to give a total of 15

(subtraction combo cannot extract a 14/2 from it) -> r5c5=(123)

Now r4c4 cannot be 2|3, otherwise we'd have a naked triple {123} in r46c4+r5c5, requiring r5c4+r6c5=18, impossible!

-> r4c34=[47|65|74]

Next, we see that there is a strong link of 5 in r8c17, so r4c1=r5c7 cannot both be 5

-> r4c1=r5c7=(478), r4c7=(347)

4. Take a look at r56c6=10:

Case 1: r56c6=[19|28|37] -> naked triple {123} in n5 -> r4c56=11={47|56} -> naked quad of {4567} in r4c3456 -> r45c7=[38]

Case 2: r56c6={46} -> r8c67=[27] -> r45c7=[38]

So no matter what r45c7=[38]!

-> r45c7=[38], r4c1=8, HS r8c2=8

5. Notice that on c1, one of r569c1 must be 4. So r6c3 and r7c2 cannot both be 4, one of them must be 5.

Case 1: r6c3=5 -> r6c4=2 -> r4c56 cannot be {29}

Case 2: r7c2=5 -> r8c1=7 -> r8c67=[45] -> r56c6 cannot be {46} -> r5c6=(123) -> naked triple in n5 -> r4c56 cannot be {29}

So no matter what r4c56 cannot be {29}!

-> r4c56={47|56}, naked quad on r4c3456, r4c89={29}, r56c6=[19|28|37] (hidden triple {123} in n5), 5 in n5 locked in r4c456

6. Since r456c4>=15, if r5c4=7 then r456c4=[573], but that would conflict with r4c56={47|56}

-> r5c4=9 -> HS r6c1=r9c2=9 -> r56c6=[28|37] -> HS r5c5=1 -> n1 contains a naked pair {67} in r12c2

7. {24} on c7 is locked in r123c7 -> r13c6+r2c8 must contain {79} (the 3 11/2 cages) -> r2c456 cannot have {79}

Also, r1c4=7 would force r2c2=7 and rule out any 7 in r13c6+r2c8 -> r1c4<>7

Hence r1c34=11=[83|92] (-> naked pair {23} on r16c4)

One of r13c6 must be 9 -> one of r13c7 must be 2 -> r2c7<>2

8. From step 5 above we know that either r6c3 or r7c2 must be 5:

Case 1: r6c3=5 -> r6c4=2 -> r1c4<>2

Case 2: r7c2=5 -> r3c12=[24] -> r3c7<>2 -> r1c7=2 -> r1c4<>2

So no matter what r1c4<>2!

-> r1c34=[83]

And the rest is all trivial!

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

Pyrrhon wrote:I decided to make a second inequality killer. I hope you have fun.

I had fun to solve it, thanks

I liked the idea of fixing the various sums steps by step after solving some parts of the puzzle.

- Jean-Christophe
**Posts:**149**Joined:**22 January 2006

For the 2nd one, I fixed most of the cages almost at the starting stage... Here is a brief starting walkthrough:

r67c9 = r78c8 > r34c8 = r5c89 > r4c67 = r56c6 > r7c456 = r8c67 = r9c34 = r9c567 > r78c23 > r4c1+r456c2 >= 10

r4c1+r456c2 = 10..12 = {123456}

r78c23 = 11..13 = {1234567}

r7c456 = r8c67 = r9c34 = r9c567 = 12..14

r4c67 = r56c6 = 13..15 = {456789}

r34c8 = r5c89 = 14..16 = {56789}

r67c9 = r78c8 = 15..17 = {6789}

If r67c9 = r78c8 = 17 = {89}, then r34c8 (14..16) cannot be larger than {67} = 13

If r67c9 = r78c8 = 16 = {79}, then r34c8 must be {68} = 14, and r5c89 (14..16) cannot be larger than [58] = 13

Hence:

r67c9 = r78c8 = 15 = {69|78}

r34c8 = r5c89 = 14 = {59|68}

r4c67 = r56c6 = 13 = {49|58|67}

r7c456 = r9c567 = 12, r8c67 = r9c34 = 12 = {39|48|57}

r78c23 = 11 = {1235}

r4c1+r456c2 = 10 = {1234}

-> r78c8={78}, r34c8=[95], r5c89=[68], r67c9=[96], r1c89=[47] (>10)...

Definitely not as difficult as the first...

BTW Uwe, I see you've uploaded another batch... Is it possible to distinguish those updated in August and those updated in September?

r67c9 = r78c8 > r34c8 = r5c89 > r4c67 = r56c6 > r7c456 = r8c67 = r9c34 = r9c567 > r78c23 > r4c1+r456c2 >= 10

r4c1+r456c2 = 10..12 = {123456}

r78c23 = 11..13 = {1234567}

r7c456 = r8c67 = r9c34 = r9c567 = 12..14

r4c67 = r56c6 = 13..15 = {456789}

r34c8 = r5c89 = 14..16 = {56789}

r67c9 = r78c8 = 15..17 = {6789}

If r67c9 = r78c8 = 17 = {89}, then r34c8 (14..16) cannot be larger than {67} = 13

If r67c9 = r78c8 = 16 = {79}, then r34c8 must be {68} = 14, and r5c89 (14..16) cannot be larger than [58] = 13

Hence:

r67c9 = r78c8 = 15 = {69|78}

r34c8 = r5c89 = 14 = {59|68}

r4c67 = r56c6 = 13 = {49|58|67}

r7c456 = r9c567 = 12, r8c67 = r9c34 = 12 = {39|48|57}

r78c23 = 11 = {1235}

r4c1+r456c2 = 10 = {1234}

-> r78c8={78}, r34c8=[95], r5c89=[68], r67c9=[96], r1c89=[47] (>10)...

Definitely not as difficult as the first...

BTW Uwe, I see you've uploaded another batch... Is it possible to distinguish those updated in August and those updated in September?

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

Pyrrhon

Of course a program is not necessary, however my writing is so scruffy that where the problem is more complicated I almost always make notation mistakes. Hence I like to do them on spreadsheet or on a human style program like JC's. I will admit that when I am doing them at lunch time or when I get completely stuck I will use his "Big Clue" function.

On this one did it first time without "help" and failed, tried again with JC's help and again failed (i no doubt made some mistake), and then tried again half and half and got it out. It is depressing that my marking up mistakes have now polluted to computer forms as well as to pencil.

I will have a look at your step this evening and try your new inequality puzzle - thank you.

Of course a program is not necessary, however my writing is so scruffy that where the problem is more complicated I almost always make notation mistakes. Hence I like to do them on spreadsheet or on a human style program like JC's. I will admit that when I am doing them at lunch time or when I get completely stuck I will use his "Big Clue" function.

On this one did it first time without "help" and failed, tried again with JC's help and again failed (i no doubt made some mistake), and then tried again half and half and got it out. It is depressing that my marking up mistakes have now polluted to computer forms as well as to pencil.

I will have a look at your step this evening and try your new inequality puzzle - thank you.

- HATMAN
**Posts:**205**Joined:**25 February 2006

Thanks Uwe...

Have you taken a look at my walkthrough to the original version of puzzle 1? Do you think it's logical enough?

BTW the relationship between r2c89 could be "<" or ">"... Both could produce a unique solution (I leave it at the very end of the puzzle to decide the final 6 cells)...

Have you taken a look at my walkthrough to the original version of puzzle 1? Do you think it's logical enough?

BTW the relationship between r2c89 could be "<" or ">"... Both could produce a unique solution (I leave it at the very end of the puzzle to decide the final 6 cells)...

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

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