# Eric S. Emrick

Updated: 10 hours 42 min ago

### Grapes of Math Puzzle

Thu, 2006-06-22 21:29
First, let me say that I was really excited about the title I gave to this post. It hit me while mowing my lawn and made me stop and laugh - thankfully the neighbors were not watching (as far as I know). Anyway, I did a quick Google search on the title Grapes of Math to see how original it was. While I had not heard of the title, it has many hits on the Web. Oh well, so much for absolute originality.

Tom Kyte has a really good puzzle on his blog. I enjoy a good puzzle and submitted my response, which I believe is the solution to the puzzle. But, for my personal satisfaction and to address Mr. Ed's concerns, I wanted to prove that my response was the only possible correct answer, given some conditions I have derived from Tom's post, the problem (picture) itself and intuition. If you are interested in the puzzle, please visit Tom's blog and try to solve it for yourself and skip the remainder of this post.

You mathematicians will please forgive any seemingly barbaric notations or proof layout :) Cary, Jonathan, Tom or any other mathematician lurking about, please feel free to critique the proof if it is incorrect.

Grapes of Math equation generated from picture:
(10(banana)+apple)/pear = 10(grapes)+peach+(strawberry/pear)

Prove the only solution set (banana, apple, pear, grapes, peach, lemon, strawberry) for the Grapes of Math is (9,3,2,4,6,8,1) given the following conditions:

Conditions

0) each fruit represents a distinct integer that must be in [0-9]. Negative integers don't really make much sense in this case - how do you ascribe a negative integer to any fruit but the pear?
1) the numerator concat(banana,apple) is in [00-99].
2) pear cannot be 1 because grapes * pear would equal grapes and it does not. grapes * pear = lemon.
3) pear cannot be 0 because x/0 is undefined for all integers x.
4) from 2) and 3) 9 >= pear > 1.
5) grapes cannot be 1 because grapes * pear would equal pear and it does not. grapes * pear = lemon.
6) grapes cannot be greater than 4 because that would yield a concat(banana,apple) that is > two digits, which cannot be (condition 1). For example, the integer portion of the quotient concat(grapes,peach) must be less than 50, based on 4).
7) from 5) and 6), 4 >= grapes > 1.
8) from 7) banana is in [4-9]. If the integer portion of the quotient (grapes) is 2, 3 or 4, then given 4) the numerator, concat(banana,apple), must be in [40-99] . The least the numerator could be is 40 given 4) and 7). The highest would be 99 by definition.
9) all fruits taste really yummy (This is for Mr. Ed)

Proof by Exhaustion (brute force method): Grapes of Math Puzzle

Case 1: grapes = 4

If grapes = 4 then banana can only be 8 or 9 because of 4).

Case 1.1: banana = 8

If grapes = 4 and banana = 8 then pear = 2 and lemon = 8. Lemon cannot equal banana by condition 0) and, thusly, banana != 8. Therefore, concat(banana,apple) is not in [80-89].

Case 1.2: banana = 9

If grapes = 4 and banana = 9 then pear = 2 and lemon = 8. Then by subtraction (banana - lemon ) = (9-8) = 1 = strawberry.

Case 1.2.1: apple = 0

If apple = 0 then peach = 5 and concat(grapes,peach) = 45 with no remainder. We know that there must be a remainder because strawberry is 1 in this case. Therefore, apple != 0 and concat(banana,apple) is not 90.

Case 1.2.2: apple = 1

If apple = 1 then apple = strawberry = 1. Therefore, apple != 1 and concat(banana,apple) is not 91.

Case 1.2.3: apple = 2

If apple = 2 then apple = pear = 2. Therefore apple != 2 and concat(banana,apple) is not 92.

Case 1.2.4: apple = 3

If apple = 3 then peach = 6 and concat(strawberry,apple) - concat(strawberry,pear) = strawberry = 1. Therefore, grapes = 4, banana = 9, pear = 2, lemon = 8, strawberry = 1 and apple = 3.

Therefore, concat(banana,apple) = 93 is a numerator solution.

Case 1.2.5: apple = 4

If apple = 4 then apple = grape = 4. Therefore, apple != 4 and concat(banana,apple) != 94.

Case 1.2.6: apple = 5

If apple = 5 then peach = 7 and pear = 4. But, pear is assumed to be 2 and cannot 2 != 4. Therefore, apple != 5 and concat(banana,apple) != 95.

Case 1.2.7: apple = 6

If apple = 6 then peach = lemon = 8. Therefore, apple != 6 and concat(banana,apple) != 96.

Case 1.2.8: apple = 7

If apple = 7 then peach = lemon = 8. Therefore, apple != 7 and concat(banana,apple) != 97.

Case 1.2.9: apple = 8

If apple = 8 then apple = lemon = 8. Therefore, apple != 8 and concat(banana,apple) != 98.

Case 1.2.10: apple = 9

If apple = 9 then apple = banana = 9. Therefore, apple != 9 and concat(banana,apple) != 99.

Therefore, for grapes = 4, the only solution for numerator concat(banana,apple) in [80-99] is 93.

Case 2: grapes = 3

If grapes = 3 then banana can only be 6 or 7 because pear > 1 from condition 4).

Case 2.1: banana = 6

If grapes = 3 and banana = 6 then pear = 2 and lemon = 6, and lemon = banana = 6. Therefore, banana != 6 and concat(banana,apple) is not in [60-69].

Case 2.2: banana = 7

If grapes = 3 and banana = 7 then pear = 2 and lemon = 6. This means apple can only be in [4-5] (cannot be 6 because lemon = apple = 6 violates condition 0).

Case 2.2.1: apple = 4

If apple = 4 then peach = 3, and peach = grapes = 3. Therefore, apple !=4.

Case 2.2.2: apple = 5

If apple = 5 then peach = 5. Therefore, apple != 5.

Therefore, concat(banana,apple) is not in [70-79].

Case 3: grapes = 2

If grapes = 2 then banana must be in [4-5] because of condition 4).

Case 3.1: banana = 4

If banana = 4 then lemon = banana. Therefore, banana != 4 and concat(banana,apple) is not in [40-49].

Case 3.2: banana = 5

If grapes = 2 and banana = 5 then pear = grapes = 2. Therefore, banana != 5 and concat(banana,apple) is not in [50-59].

Therefore, concat(apple,banana) is not in [40 - 59].

From grapes in [2-4], we have proved that only one solution (93) exists for the numerator concat(banana,apple) between 40 and 99. By condition 7), concat(banana,apple) is not in [00-39].

Conclusion

Therefore, 93 is the only solution for concat(banana,apple) in [00-99]. After exhausting all possible two digit values for concat(banana,apple) only one solution set (banana, apple, pear, grapes, peach, lemon, strawberry) was found:

(9,3,2,4,6,8,1)

and for Mr. Ed...

(9,3,-2,4,6,8,1) iff concat(grapes,peach) = -46

Solution set applied to equation of Grapes of Math:

(10(9)+3)/2=93/2=46 ½=10(4)+6+(1/2) quod erat demonstrandum

Note: lemon is absorbed in the equation, given the correctness of strawberry = 1.

### Oracle Blooper

Wed, 2006-06-21 20:21
Well, instead of assimilating the suitable, albeit plentiful, Oracle manuals to cover the material required for the Oracle 7.3 -> Oracle 9i upgrade certification exam I opted to search for a single resource. I came across a third-party study tool that was recommended by Oracle and seemed to foot the bill. Very much to my surprise the material was disjointed, loaded with typographical errors and often flat out wrong.

I was willing to forgive the copious typographical mistakes per/page (which approached the Golden Ratio mind you) as I could deduce the intentions. I could stomach the disjointed word salads and sparse information. But I refused to read another page after encountering a heinously blatant, careless and nonsensical bit of misinformation. How could I possibly continue to use this material as a study reference if I could not trust the content? To the misinformation at hand, the material states the following verbatim:

PGA_USED_MEM - The process is using PGA memory.
PGA_ALLOC_MEM- The process has been allocated PGA memory.
PGA_MAX_MEM - The process has been allocated maximum memory.
PGA_GIBBERISH - The process has found gibberish in the PGA and wishes to purge. (OK, this was my invention)

I scratched my head. Re-read, scratched head some more. Finished beer and reached for another. Nothing seemed to alleviate my consternation. I was well aware of these attributes of v\$process and was not so much concerned with the incorrectness, as I knew their meaning. It was the gross negligence that left my jaw drooping for a minute.

The values for these attributes are NOT Boolean as you well know. You don't query v\$process and find a Y or N associated with the values for these attributes. The Oracle documentation defines these attributes in a very straightforward manner. Is there any other way?

PGA_USED_MEM number PGA memory currently used by the process
PGA_ALLOC_MEM number PGA memory currently allocated by the process (including free PGA memory not yet released to the operating system by the server process)
PGA_MAX_MEM number Maximum PGA memory ever allocated by the process

Simply stated, I was shocked that the author(s) and editor(s) put such little thought into the material and subsequent proof reading. Actually, I think the author's brain was tied behind his back while writing this material. If one aspires to put together training material and includes attribute definitions that are pre-defined for you in the Oracle documentation set, might I recommend taking a cursory glance at said documentation? You can't just feed me a heaping helping of documentation rubbish without expecting me to pitch the kindling into the nearest can - I know, I've seen me do it! Did I mention the material is several fold more expensive than any of Tom Kyte's or Jonathan Lewis' books? Lesson learned.

### The Oracle Certification Process

Sun, 2006-04-30 20:13

In late 1997, on what was a very shiny day (for all of DBA-kind I am sure), I proudly exited my local facility that proctored Oracle certification exams. On this glorious day I had passed the last of four exams required to obtain the coveted Oracle Certified Professional (OCP) title. I was certified on Oracle 7.3 and could not have been more proud. After waiting a few weeks to receive my certificate I brandished it in my home study. Make no mistake about it. I felt this had legitimized my 3 long years of Oracle work to date - I had reached an Oracle summit. At that time the OCP title was not nearly as pervasive as it is today. In hindsight, I suppose my enthusiasm was not entirely unjustified.

Let’s roll time forward nine years to 2006. I have not renewed my certification. For all practical purposes I am not an OCP. I certainly wouldn’t claim such on my resume having only achieved version 7.3 certification. Why haven’t I renewed my certification? After all, Oracle has bent over backwards to assist this erstwhile OCP by offering an upgrade exam. I can take a single exam and immediately upgrade my certification status to an Oracle 9i OCP. If I labored a bit more, I could take another upgrade exam and attain the highest OCP level available. Does this mean that I could, nearly overnight, claim expertise in all of the concepts and elegant nuances Oracle has built into its database since version 7.3? Professionally, on my resume, I suppose the answer is yes. Realistically, the answer is, no way!

I feel the only real way to stay current with our Oracle knowledge and exhibit the technical acumen associated with a proficient Oracle practitioner is to read (and reread) documentation and test features. There is absolutely no substitute for good old-fashioned studying in conjunction with trial and error exercises. I have interviewed dozens of Oracle Certified Professionals over the years, many of which struggled with the basics. I do believe that today, more than ever, the ubiquitous OCP title provides little insight into the qualifications of an Oracle DBA. However, I do believe that the certification process can lay an excellent framework for a strong understanding of the Oracle database. Just, not by necessity. It varies from person to person. One person with the same temporal experience with Oracle and an OCP title might appear lacking when compared to another with equal “qualifications” and accomplishments. Why? We all have different approaches to storing information for retrieval. I remember cramming for exams in college for the courses I loathed. I always seemed to make out okay. But, did I really learn the material or just buffer it long enough so that my mind could hurl it back out in the nick of time? I know, for those “undesirable” classes it was the latter. For me to learn I must:

1. Want to learn.
2. Be passionate about the topic.
3. and study, study, study.

Of course, there are exceptions to the rules, those supremely intelligent humans that roam the earth with a glut of gray matter that have little need for 3), leaving it for the rest of us to toil.

Am I a better DBA than I was nine years ago? I certainly hope so. Could I augment the breadth and depth of my Oracle knowledge by revisiting the certification process? Absolutely. But, couldn’t I really do the same by studying the material covered by the exams? After all, I am passionate about the topic and want to learn. I know. I know. It sounds like a really cheap excuse. Read the material, but, uh hum, skip the exams right? How convenient.

For those of you with your OCP please don’t think I am minimizing your achievements. I am certainly not doing so. I believe that the Oracle certification process can yield a very productive learning experience, insofar as we really take the time to authentically learn the material we are studying. It has been my experience, that if I have ostensibly forgotten what I have learned, as long as I truly understood the material while in the learning process, re-learning can be a very quick enterprise.

By the way, I think I will take the upgrade exams this year. But, this time I refuse to cram. I will revisit the exam topics with a cheerful willingness, as the science of Oracle database administration is a very exciting and challenging branch of knowledge.

### Solaris and High Wait I/O CPU

Tue, 2006-04-25 17:11
A few days back a familiar little situation surfaced. Someone monitoring the OS was making claims that a particular machine was running at 100% cpu utilization during a period when a portion of the application was running slower than normal. The assertion being made was our system had a cpu shortage.

Given the fact that the application was running on a Solaris platform I looked at the vmstat history logs kept for just such an investigation. Per vmstat, for the time in question, there was plenty of idle cpu. Immediately, I thought this person must have been looking at the sar data on the machine in question. Sure enough, the sar data indicated a very low percentage of idle cpu. As you might have guessed, the percentage of time the system was waiting for I/O was rather large according to sar and, consequently, low idle time was being reported. I explained that it was typical for this system to run a high wait I/O percentage as reported by sar; after all, it is a database server with many processors. I also explained that low idle time as reported by sar does not necessarily mean a cpu bottleneck exists.

I remembered reading in Adrian Cockroft’s book, Sun Performance Tuning, that vmstat lumps wait I/O into idle time. So, naturally I was confident in my counter-assertion that our cpu utilization was just fine. I assuredly reached for my copy of the Sun Performance Tuning book to show where I had read this information years ago. I searched the index of the book and gave the book a cursory once-over to no avail. I started doubting whether I had reached for the wrong text! A bit frustrated I decide to perform a full book scan. Low and behold, I only got past two pages before my memory was vindicated. On page 3 it reads. “Whenever there are any blocked processes, all cpu idle time is treated as wait for I/O time! The vmstat command correctly includes wait for I/O in its idle value…” Viola!

The clock interrupt handler in the Solaris operating system runs every 10ms (or at least used to) to get cpu utilization information. It will search the state structure for each cpu and find that each cpu is in one of five states: user, system, idle, waiting for I/O or quiesced. Based on my understanding, the quiesced state is not really indicated by a value stored in a structure or variable associated with a cpu. It is simply the state when a cpu is not running user, system or idle threads and not waiting for I/O.

The point is, a high value for wait I/O generated from sar on a Solaris platform does not indicate a cpu bottleneck. Moreover, high wait I/O values do not necessarily indicate an I/O bottleneck. However, an I/O bottleneck could very easy manifest in high wait I/O percentages. You really need to look at your I/O service times to determine if the I/O subsystem is performing poorly.

For those wanting to know more on the algorithm used by Solaris to calculate idle and wait I/O cpu percentages read here. It is a bit dated, but describes how wait I/O is tallied in the Solaris operating system (at least in earlier versions). Interestingly enough this article cites Sun Performance Tuning, my trusty reference.

### Oracle Riddles: Now that is interesting.

Wed, 2006-03-29 21:11
A process can prevent me from staking the claim for another. Users sometimes need me to know if they can secure a spot where another process may have been before. Needless to say, I am a pretty big deal, and users are very interested in me. However, when some stop by for a visit they often feel I haven't cleaned up very well. Do you know what I am?