View previous topic :: View next topic 
Author 
Message 
keith
Joined: 19 Sep 2005 Posts: 3253 Location: near Detroit, Michigan, USA

Posted: Sat Dec 17, 2005 9:41 pm Post subject: A very tough but solvable puzzle 


Here is a very tough puzzle, actually published (it is David Bodycombe's puzzle of 10 Dec 2005, published locally in the Detroit Free Press).
Code: 
++++
 . . .  6 . 2  . 8 . 
 . . .  . . .  . 1 5 
 4 . 3  . 1 .  . . . 
++++
 8 . .  . . 7  . . 4 
 . . 1  . 2 .  6 . . 
 7 . .  9 . .  . . 2 
++++
 . . .  . 3 .  8 . 7 
 1 6 .  . . .  . . . 
 . 7 .  4 . 8  . . . 
++++

Although I can tell you how a human might solve it (without Nishio or guessing), it is way beyond my own capability
Keith
Michigan, USA 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3253 Location: near Detroit, Michigan, USA

Posted: Sat Dec 17, 2005 9:49 pm Post subject: Apology 


I now see this one was discussed on Monday.
My solution involves, in addition to the usual,
XY Wing
Forcing Chain
Forcing Chain
"Jellyfish" (a complicated Swordfish)
XY Wing
Forcing Chain
Does anyone still have an interest in this one?
Keith 

Back to top 


David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Sun Dec 18, 2005 12:50 am Post subject: I'm interested 


Keith wrote:  Does anyone still have an interest in this one? 
I'm interested. I didn't spot a jellyfish in this puzzle, and I'd like to see that.
Do you get these puzzles (Detroit Free Press) regularly? I only get this particular series of puzzles on Saturdays, and lately it's been a pretty tough puzzle every Saturday. I particularly liked the one for today (12/17). Have you looked at that one?
Happy sudokuing! dcb 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3253 Location: near Detroit, Michigan, USA

Posted: Sun Dec 18, 2005 6:21 am Post subject: Will post solution 


I'll post a solution after I have written it up.
This is a daily puzzle. Recently, the Saturday ones have been graded with six stars, and contain X Wings, XY Wings, etc. The other daily puzzles require at most a couple of block  column interactions, along with pinned squares, twins, triplets, etc.
I have not seen that these puzzles ar published or archived on a web site. I suppose you could find them at a newspaper site. The one on the Detroit Free Press site (www.freep.com, look under "comics") does not seem to be the same as the one printed in the paper.
The one for today was tough for me, as I recall it had an XYWing. It is MUCH easier than this one.
Keith 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3253 Location: near Detroit, Michigan, USA

Posted: Sun Dec 18, 2005 7:10 am Post subject: Detailed solution for this tough puzzle 


Here is the solution, as given by Sudoku Susser
http://www.madoverlord.com/projects/sudoku.t
This is an interesting solver, because it solves puzzles the way a human would. Also, you can turn various heuristics on and off. For example, in this puzzle, if you turn the jellyfish off, it resorts to Nishio, which implies that the jellyfish is essential  if you don't find it, you have to use trial and error.
The program takes a long time to find the last forcing chain in this puzzle.
The program also has a very good manual which explains all the patterns, strategies, and heuristics for solving Sudoku. I normally solve puzzles with pencil and paper (including the Daily Sudoku, every day!), and resort to the program only as a learning tool. Then, it is very good. Take a look at the explanation of the Unique Rectangle given below. Even I can understand it!
Best wishes,
Keith
===========================
* R1C2 is the only square in row 1 that can be <1>.
* R5C8 is the only square in row 5 that can be <7>.
* R6C5 is the only square in row 6 that can be <8>.
* R5C9 is the only square in row 5 that can be <8>.
* R8C3 is the only square in row 8 that can be <8>.
* R2C1 is the only square in column 1 that can be <6>.
* R9C9 is the only square in column 9 that can be <1>.
* R3C9 is the only square in column 9 that can be <6>.
* R9C1 is the only square in block 7 that can be <3>.
* R7C1 is the only square in column 1 that can be <2>.
* R8C4 is the only square in column 4 that can be <2>.
* R8C5 is the only square in row 8 that can be <7>.
* Squares R3C6 and R8C6 in column 6 form a simple naked pair. These 2 squares both contain the 2 possibilities <59>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the column.
R2C6  removing <9> from <349> leaving <34>.
R5C6  removing <5> from <345> leaving <34>.
R6C6  removing <5> from <13456> leaving <1346>.
R7C6  removing <59> from <1569> leaving <16>.
* Squares R2C6 and R5C6 in column 6 form a simple naked pair. These 2 squares both contain the 2 possibilities <34>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the column.
R6C6  removing <34> from <1346> leaving <16>.
* R5C6 is the only square in block 5 that can be <4>.
From this deduction, the following moves are immediately forced:
R2C6 must be <3>.
* A set of 2 squares form a simple hidden pair. R2C4 and R3C4 all contain all of the 2 possibilities <78>. No other squares in column 4 have those possibilities. Since the 2 squares are the only possible locations for 2 possible values, any additional possibilities these squares have (if any) can be eliminated. These squares now become a simple naked pair.
R3C4  removing <5> from <578> leaving <78>.
* Intersection of row 5 with block 4. The value <9> only appears in one or more of squares R5C1, R5C2 and R5C3 of row 5. These squares are the ones that intersect with block 4. Thus, the other (nonintersecting) squares of block 4 cannot contain this value.
R4C2  removing <9> from <2359> leaving <235>.
R4C3  removing <9> from <2569> leaving <256>.
* Intersection of row 8 with block 9. The values <34> only appears in one or more of squares R8C7, R8C8 and R8C9 of row 8. These squares are the ones that intersect with block 9. Thus, the other (nonintersecting) squares of block 9 cannot contain these values.
R7C8  removing <4> from <4569> leaving <569>.
* R8C8 is the only square in column 8 that can be <4>.
* Intersection of column 8 with block 6. The value <3> only appears in one or more of squares R4C8, R5C8 and R6C8 of column 8. These squares are the ones that intersect with block 6. Thus, the other (nonintersecting) squares of block 6 cannot contain this value.
R4C7  removing <3> from <1359> leaving <159>.
R6C7  removing <3> from <135> leaving <15>.
* Squares R1C9, R8C9, R1C7 and R8C7 form a Type4B Unique Rectangle on <39>. Because they share two rows, two columns, and two blocks, if they all had possibilities <39> then the puzzle would have two solutions; you could simply exchange the <3>'s with the <3>'s in the squares to get the other solution, and their common rows, columns and blocks would still contain one of each value. If you look carefully, you'll see that the only squares in column 7 that can contain <3> are the "roof" squares  R1C7 and R8C7. Since one of these squares must be <3>, the only way to avoid the "deadly pattern" is if neither of them can contain <9>.
R1C7  removing <9> from <3479> leaving <347>.
R8C7  removing <9> from <359> leaving <35>.
* Squares R6C6 (XY), R6C7 (XZ) and R4C5 (YZ) form an XYWing pattern on <5>. All squares that are buddies of both the XZ and YZ squares cannot be <5>.
R4C7  removing <5> from <159> leaving <19>.
R4C8  removing <5> from <359> leaving <39>.
* Intersection of block 6 with row 6. The value <5> only appears in one or more of squares R6C7, R6C8 and R6C9 of block 6. These squares are the ones that intersect with row 6. Thus, the other (nonintersecting) squares of row 6 cannot contain this value.
R6C2  removing <5> from <345> leaving <34>.
R6C3  removing <5> from <456> leaving <46>.
* Found a forcing chain. If we assume that square R3C8 is <9> then we can make the following chain of conclusions:
R3C6 must be <5>, which means that
R8C6 must be <9>, which means that
R8C9 must be <3>, which means that
R1C9 must be <9>, which means that
R3C8 must be <2>.
Since this is logically inconsistent, R3C8 cannot be <9>.
(772 links were considered before finding this chain)
* R9C7 is the only square in row 9 that can be <2>.
* Found a forcing chain. If we assume that square R3C7 is <9> then we can make the following chain of conclusions:
R3C6 must be <5>, which means that
R8C6 must be <9>, which means that
R8C9 must be <3>, which means that
R1C9 must be <9>, which means that
R3C7 must be <7>.
Since this is logically inconsistent, R3C7 cannot be <9>.
(982 links were considered before finding this chain)
From this deduction, the following moves are immediately forced:
R3C4 must be <8>.
R2C4 must be <7>.
* R1C3 is the only square in row 1 that can be <7>.
* R2C2 is the only square in row 2 that can be <8>.
* R2C3 is the only square in row 2 that can be <2>.
* R4C2 is the only square in row 4 that can be <2>.
* Squares R4C3 and R4C5 in row 4 form a simple naked pair. These 2 squares both contain the 2 possibilities <56>. Since each of the squares must contain one of the possibilities, they can be eliminated from the other squares in the row.
R4C4  removing <5> from <135> leaving <13>.
* Intersection of column 3 with block 7. The value <9> only appears in one or more of squares R7C3, R8C3 and R9C3 of column 3. These squares are the ones that intersect with block 7. Thus, the other (nonintersecting) squares of block 7 cannot contain this value.
R7C2  removing <9> from <459> leaving <45>.
* Squares R2C5 and R2C7 in row 2, R4C7 and R4C8 in row 4, R7C3 and R7C8 in row 7 and R9C3, R9C5 and R9C8 in row 9 form a Jellyfish pattern on possibility <9>. All other instances of this possibility in columns 3, 5, 7 and 8 can be removed.
R1C5  removing <9> from <459> leaving <45>.
* Squares R6C2 (XY), R6C8 (XZ) and R7C2 (YZ) form an XYWing pattern on <5>. All squares that are buddies of both the XZ and YZ squares cannot be <5>.
R7C8  removing <5> from <569> leaving <69>.
* Found a forcing chain. If we assume that square R1C5 is <5> then we can make the following chain of conclusions:
R1C1 must be <9>, which means that
R5C1 must be <5>, which means that
R4C3 must be <6>, which means that
R4C5 must be <5>, which means that
R1C5 must be <4>.
Since this is logically inconsistent, R1C5 cannot be <5>.
(13641 links were considered before finding this chain)
From this deduction, the following moves are immediately forced:
R1C7 must be <3>.
R2C5 must be <9>.
R1C9 must be <9>.
R8C7 must be <5>.
R1C1 must be <5>.
R8C9 must be <3>.
R2C7 must be <4>.
R3C6 must be <5>.
R3C2 must be <9>.
R8C6 must be <9>.
R6C7 must be <1>.
R5C1 must be <9>.
R6C6 must be <6>.
R4C7 must be <9>.
R4C8 must be <3>.
R4C4 must be <1>.
R6C8 must be <5>.
R6C3 must be <4>.
R7C6 must be <1>.
R4C5 must be <5>.
R7C4 must be <5>.
R4C3 must be <6>.
R9C5 must be <6>.
R5C4 must be <3>.
R5C2 must be <5>.
R6C2 must be <3>.
R7C2 must be <4>.
R7C3 must be <9>.
R9C8 must be <9>.
R9C3 must be <5>.
R7C8 must be <6>.
(Scientists say that mind exercises like these help prevent dementia later in life. They don't mention the danger of shortterm insanity! KM) 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3253 Location: near Detroit, Michigan, USA

Posted: Sun Dec 18, 2005 7:29 am Post subject: Today's (12/17) Bodycombe puzzle 


dcb wrote:
>>
I particularly liked the one for today (12/17). Have you looked at that one?
<<
Yes. As I recall, it came down to this unique rectangle:
25 ... 25
25 ... 257
(These are possible values for four squares at the corners of a rectangle in the puzzle.)
and these were the only occurrences of 2 in each row and column. Think about it. You cannot have
25 .. 25
25 .. 25
as the corners of a rectangle, because then, in these four squares, there are two solutions. Pick one, The other is obtained by interchanging 2 and 5. So, the bottom right square must be a 7.
If anyone is interested, I'll post this puzzle as a new thread.
Keith 

Back to top 


someone_somewhere
Joined: 07 Aug 2005 Posts: 275 Location: Munich

Posted: Sun Dec 18, 2005 9:26 am Post subject: 


Hi Keith,
I could not spot the jellyfish, probable because it is not needed.
Did not know the TYPE4B UNIQUE RECTANGLE technique.
Thank you, I learned something new.
BTW, in the solution posted (by SudokuSusser) the TYPE4B UNIQUE RECTANGLE is not a must. It can be solved also without it.
But, as you said, it is very nice and I like it.
Now, caming back to the essential:
When we as human look for "forcing chains" (I am using the similar technique, called "double implication chains") the most impoartant question is: WHERE DO WE START TO LOOK?
For a computer program this question is not so important because he can/will scan seqeuntially all the posibilities, if the programmer did not put some priorities, .... which is the same as for us ....
My approch is: "to start from the pairs" and if there are only a few and they do not bring anything, than start with the "relevant" digits of the multidigit cells. Relevant meaning for example one that appears only 2 times in a row/column/3x3 block.
As an example I have excluded 5 from r5c1 (the last forcing chain described in the solution of SudokuSusser) instead of the 5 from c1r5 as the (stupid) program was doing. And this because I started from pairs and r5c1 had the pair of digits 5 & 9. The computer started from r1c5 which was holding as candidates: 4, 5 & 9. So in my case I could mark the other digit as sure, but the computer was left with 2 digits after eliminating one digit.
(This does not mean that the algorithm can't be improved to get closer to the human logic).
Thank you for your contribution. I have learned something new.
see u, 

Back to top 


David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Sun Dec 18, 2005 9:51 pm Post subject: A uniquitous sudoku method ... 


Keith wrote:  Squares R1C9, R8C9, R1C7 and R8C7 form a Type4B Unique Rectangle on <39>. Because they share two rows, two columns, and two blocks, if they all had possibilities <39> then the puzzle would have two solutions; you could simply exchange the <3>'s with the <9>'s in the squares to get the other solution, and their common rows, columns and blocks would still contain one of each value. If you look carefully, you'll see that the only squares in column 7 that can contain <3> are the "roof" squares  R1C7 and R8C7. Since one of these squares must be <3>, the only way to avoid the "deadly pattern" is if neither of them can contain <9>. 
This is the kind of reasoning that Glassman has dubbed "uniquity." There's an interesting thread on that topic in this forum.
I agree that this is an interesting logical step. However, it is only valid if the puzzle has a unique solution. So this form of logic is essentially circular  if it is applied to a puzzle that does not have a unique solution, it may lead to one possible solution while ruling out another one (or maybe two, or three ...). In terms of classical logic, this is called assuming the consequent.
In short, for this method to be logically valid, one needs an independent means of verifying that the solution is indeed unique. dcb 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3253 Location: near Detroit, Michigan, USA

Posted: Mon Dec 19, 2005 9:13 am Post subject: Assuming the Consequent: Not at all! 


David,
A very interesting point, with which I disagree. First, a quote from WikiPedia:
"It is possible to set starting grids with more than one solution and to set grids with no solution, but such are not considered proper Sudoku puzzles; like most other purelogic puzzles, a unique solution is expected."
http://en.wikipedia.org/wiki/Sudoku
As I understand it, "Assuming the consequent" means making an assumption that leads to a particular solution. If the puzzle has only one solution, there is no logical flaw, and no circular logic, in using techniques that take advantage of this fact.
In other words, I think the statement that a puzzle has a unique solution is a ground rule, not an assumption on my part. Glassman's characterization of solution techniques that presume a unique solution as "less than elegant" is unwarranted.
Keith 

Back to top 


David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Mon Dec 19, 2005 9:28 pm Post subject: We'll have to agree to disagree ... 


Keith wrote:  If the puzzle has only one solution, there is no logical flaw, and no circular logic, in using techniques that take advantage of this fact. 
Thank you for making my point, Keith  you said "If the puzzle has only one solution ..." I, for one, can't be certain of this simply by examining the starting grid. I might be willing to take it on faith, based on the author's reputation. But that's not logical necessity. And even the most careful authors have been known to make mistakes.
Solving the puzzle logically (without assuming uniquity) proves that the solution is unique. Making the assumption that the solution is unique may indeed allow one to obtain a "solution," without demonstrating that additional "solutions" are impossible. I'll see if I can cook up an example sometime. dcb 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3253 Location: near Detroit, Michigan, USA

Posted: Wed Dec 21, 2005 2:00 am Post subject: No, we just have to agree 


David,
I think we just have to agree: Is the uniqueness of the puzzle known a priori or not?
If I were in your shoes, where (I believe) uniqueness is required but needs to be checked, I would use a computer program.
I spent some time today trying to construct a puzzle with not one, but only two solutions. It was a lot harder than I thought it would be. I am now thinking that if a puzzle has multiple solutions (and if all the numbers occur in the starting grid), then it will have many possible solutions (more than two or three).
I can think of one way to possibly make a puzzle that has two solutions, but it would be obvious from the start. Take a vaild filled grid. Remove all the 9's, and all but one of the 8's. Then, find a valid starting grid that includes the single 8, and no 9's. Delete the 8. This may be a puzzle with two solutions  the 8's and 9's can be interchanged. But, since both 8 and 9 are missing from the starting grid, it is obviously so.
Keith 

Back to top 


David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Wed Dec 21, 2005 4:39 pm Post subject: Multiple solutions 


Keith wrote:  I am now thinking that if a puzzle has multiple solutions (and if all the numbers occur in the starting grid), then it will have many possible solutions (more than two or three). 
Hi, Keith!
I've given the subject of multiple solutions a bit of thought. I think the easiest way to visualize them is in terms of the (almost) completed grid.
The very simplest case is the one the "Sudoku Susser" calls a Type4B Unique Rectangle. We just have two squares in each of two different 3x3 boxes that fall on the same columns and rows (at r1c1, r1c3, r9c1, r9c3, for instance). We can transform one "solution" into the other by simply reversing the order of the pair in both 3x3 boxes.
Code:  4 . 5 . . . . . . 5 . 4 . . . . . . (r1)
                 
                 
5 . 4 . . . . . . 4 . 5 . . . . . . (r9) 
It's pretty easy to see that we could also have an interchangeable triplet of values that fall in just two different 3x3 boxes (at r1c1, r1c2, r1c3, r9c1, r9c2, r9c3, for instance). Now there will be six different "solutions," because a set of three distinct symbols can be permuted into six (=3!) different ordered sequences.
I can also visualize a situation in which four "solutions" are possible  just set up two independent "Type4B Unique Rectangles." We ought to be able to proceed on up to grids with 8 and 16 "solutions" by setting up three and then four of these rectangles  after that we get in trouble because we can't have ten nonintersecting columns or rows.
We can also perform this construction with the triplets I mentioned before to arrive at 36 or 216 "solutions"  this could get tricky, though, because 3 sets of triplets would exhaust the set of 9 symbols, assuming they were mutually exclusive, and once the same symbol gets used in two sets of triplets we might run into problems completing the grid (ie, there might not be any possible solutions after placing 4 sets of triplets in 8 boxes).
The situation starts to get very confusing once we have permutable symbols in three or more boxes and all of them interact with each other. We could have a "hexagon" (at r1c1, r1c2, r4c1, r4c2, r7c1, r7c2, for instance) that just involves three symbols (say {1, 2} in the first row, {2, 3} in the fourth row, and {1, 3} in the seventh row). This case is pretty simple  there are only two ways it can be put together. I haven't really analyzed cases more complex than this  if you're not afraid of the temporary insanity, you might have fun with patterns involving say three different values in four different boxes. dcb 

Back to top 


David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Sun Feb 05, 2006 1:53 am Post subject: An example of "uniquity" going awry 


David Bryant wrote:  Solving the puzzle logically (without assuming uniquity) proves that the solution is unique. Making the assumption that the solution is unique may indeed allow one to obtain a "solution," without demonstrating that additional "solutions" are impossible. I'll see if I can cook up an example sometime. 
Quite by accident I managed to cook up an example  by misplacing a digit in a puzzle posted in this forum. Here's the puzzle ... besides the "2" that I misplaced by accident, I had to insert one additional digit to make it work.
Code:  . . . . 1 . . 4 8
9 . . . . . . 2 .
. 5 3 7 . . . . .
. . . . . 5 9 6 .
3 . . . 9 . . . 1
. 8 9 4 . . . . .
. 3 . . . 6 8 5 .
. . 2 . . . . . 6
5 4 . . 7 . . . . 
You can get 70 cells filled in without too much trouble. At that point you'll spot a "nonunique rectangle." If you assume that the solution is unique you'll easily complete the puzzle. And, if you assume the solution is _not_ unique you will find two more ways to finish it up. dcb 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3253 Location: near Detroit, Michigan, USA

Posted: Wed Apr 26, 2006 1:14 am Post subject: 


David,
This is the possible example where you cannot use Uniqueness on an initial square.
Keith 

Back to top 


David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Wed Apr 26, 2006 3:56 pm Post subject: I don't think so 


keith wrote:  This is the possible example where you cannot use Uniqueness on an initial square. 
I don't think that's quite right, Keith.
Here's the position reached with 70 cells completed.
Code:  6 2 7 35 1 9 35 4 8
9 1 8 6 45 34 357 2 57
4 5 3 7 8 2 6 1 9
1 7 4 8 3 5 9 6 2
3 6 5 2 9 7 4 8 1
2 8 9 4 6 1 57 3 57
7 3 1 9 2 6 8 5 4
8 9 2 35 45 34 1 7 6
5 4 6 1 7 8 2 9 3 
As you can see, the "nonunique rectangle" in r2c7&9 / r6c7&9 does not involve any "initial" squares. It looks like a perfectly good example of a nonunique rectangle. Assuming the solution is unique, one simply sets r2c7 = 3 and completes the puzzle without difficulty.
The thing is, if you set r1c7 = 3 you can also complete all the cells in the puzzle consistently except for the presence of the "deadly pattern" in r2c7&9 / r6c7&9. So the puzzle has 3 possible "solutions." dcb 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5435 Location: Rochester, NY, USA

Posted: Wed Apr 26, 2006 5:22 pm Post subject: 


David, how do you define a "nonunique rectangle"? When I see any rectangle with a common pair in each corner, I view it as potential to use the technique I know as "unique rectangle." 

Back to top 


David Bryant
Joined: 29 Jul 2005 Posts: 559 Location: Denver, Colorado

Posted: Wed Apr 26, 2006 10:09 pm Post subject: It's the same thing 


Hi, Marty!
It's the same thing. I know that Robert Woodhead calls it a Unique Rectangle. The thing is, the distinguishing feature of such a rectangle (no matter which of the many "types" it is) is that completing it in a certain way will lead to nonunique solutions. So it seems natural to me to think of it as a "nonunique rectangle."
I hope that's not too confusing. I'll try to use the quote marks every time, to alert you (and maybe others?) to the fact that this is my own little idiosyncracy. dcb
PS I ran into an example of a "nonunique hexagon" the other day, which is interesting because it's somewhat different  you can read about that in the "Nightmare" forum. 

Back to top 


keith
Joined: 19 Sep 2005 Posts: 3253 Location: near Detroit, Michigan, USA

Posted: Wed Apr 26, 2006 11:58 pm Post subject: 


I agree with David. The logic rests on avoiding solutions or possibilities that are not unique.
Also, if you are not a computer programmer, the various classifications of (non) unique rectangles are not important. Just recognize the pattern, and learn what to do with it.
Keith 

Back to top 


Marty R.
Joined: 12 Feb 2006 Posts: 5435 Location: Rochester, NY, USA

Posted: Thu Apr 27, 2006 5:36 pm Post subject: 


Thanks David.
Keith, I'm not a computer programer, although I spent nine years as one in the dark ages of mainframe computers which occupied an entire large room and whose tape drives were the size of refrigerators.
I have looked at the rectangle classificationsI assume you mean Type 1, 2, etc.and have no idea of what I can do with the knowledge that a rectangle is Type "X." 

Back to top 




You cannot post new topics in this forum You cannot reply to topics in this forum You cannot edit your posts in this forum You cannot delete your posts in this forum You cannot vote in polls in this forum

Powered by phpBB © 2001, 2005 phpBB Group
