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1. Shard Calculation

Query:

Response:

Calculated Shard: 32 (from laksa063)

2. Crawled Status Check

Query:

curl -X POST \
  'http://laksa032.int.ahrefs:8124/' \
  -H 'Content-Type: text/plain' \
  -H 'X-ClickHouse-Database: crawler3' \
  -H 'Authorization: Basic YXBpOg==' \
  -d 'SELECT getAhrefsURLFromUnparsed(src_unparsed) AS found_url, ifNull(toUnixTimestamp(download_stamp), 0) AS crawl_time, ifNull(toUnixTimestamp(props_url_first_seen), 0) AS first_indexed_time, download_http_code AS http_code, src_unparsed AS src_unparsed, src_root_hash AS src_root_hash, history_drop_reason AS history_drop_reason, meta_title AS meta_title, meta_descriptions AS meta_descriptions, attrs_boilerpipe_text AS attrs_boilerpipe_text, attrs_markdown AS attrs_markdown, attrs_readable_markdown AS attrs_readable_markdown, meta_canonical AS meta_canonical FROM crawler3.page_info_local FINAL PREWHERE (src_root_hash, src_unparsed) IN ((getAhrefsRootHashFromUnparsed(getAhrefsUnparsedNoserviceFromURL(\'https://mca.ignougroup.com/2017/01/why-is-it-best-to-use-prime-number-as.html\')), getAhrefsUnparsedNoserviceFromURL(\'https://mca.ignougroup.com/2017/01/why-is-it-best-to-use-prime-number-as.html\'))) FORMAT JSONEachRow'

Response:

{"found_url":"https:\/\/mca.ignougroup.com\/2017\/01\/why-is-it-best-to-use-prime-number-as.html","crawl_time":1776279578,"first_indexed_time":1748786820,"http_code":200,"src_unparsed":"com,ignougroup!mca,\/2017\/01\/why-is-it-best-to-use-prime-number-as.html s443","src_root_hash":"3437017450487322632","history_drop_reason":null,"meta_title":"Why is it best to use a prime number as a mod in a hashing function? | MCA IGNOU GROUP","meta_descriptions":[],"attrs_boilerpipe_text":"Question Detail:\n \nIf I have a list of key values from 1 to 100 and I want to organize them in an array of 11 buckets, I've been taught to form a mod function\n$$ H = k \\bmod \\ 11$$\nNow all the values will be placed one after another in 9 rows. For example, in the first bucket there will be $0, 11, 22 \\dots$. In the second, there will be $1, 12, 23 \\dots$ etc.\nLet's say I decided to be a bad boy and use a non-prime as my hashing function - take 12. Using the Hashing function\n$$ H = k \\bmod \\ 12$$\nwould result in a hash table with values $0, 12, 24 \\dots $ in the first bucket, $1, 13, 25 \\dots$ etc. in the second and so on.\nEssentially they are the same thing. I didn't reduce collisions and I didn't spread things out any better by using the prime number hash code and I can't see how it is ever beneficial.\nAsked By : \nCodyBugstein\nAnswered By : \nMario Cervera\nConsider the set of keys $K=\\{0,1,...,100\\}$ and a hash table where the number of buckets is $m=12$. Since $3$ is a factor of $12$, the keys that are multiples of $3$ will be hashed to buckets that are multiples of $3$:\nKeys $\\{0,12,24,36,...\\}$ will be hashed to bucket $0$.\nKeys $\\{3,15,27,39,...\\}$ will be hashed to bucket $3$.\nKeys $\\{6,18,30,42,...\\}$ will be hashed to bucket $6$.\nKeys $\\{9,21,33,45,...\\}$ will be hashed to bucket $9$.\nIf $K$ is uniformly distributed (i.e., every key in $K$ is equally likely to occur), then the choice of $m$ is not so critical. But, what happens if $K$ is not uniformly distributed? Imagine that the keys that are most likely to occur are the multiples of $3$. In this case, all of the buckets that are not multiples of $3$ will be empty with high probability (which is really bad in terms of hash table performance).\nThis situation is more common that it may seem. Imagine, for instance, that you are keeping track of objects based on where they are stored in memory. If your computer's word size is four bytes, then you will be hashing keys that are multiples of $4$. Needless to say that choosing $m$ to be a multiple of $4$ would be a terrible choice: you would have $3m\/4$ buckets completely empty, and all of your keys colliding in the remaining $m\/4$ buckets.\nIn general:\nEvery key in $K$ that shares a common factor with the number of buckets $m$ will be hashed to a bucket that is a multiple of this factor.\nTherefore, to minimize collisions, it is important to reduce the number of common factors between $m$ and the elements of $K$. How can this be achieved? By choosing $m$ to be a number that has very few factors: a\nprime number\n.","attrs_markdown":"[World's most popular travel blog for travel bloggers.](http:\/\/circularjourney.com\/?utm_source=mca-blog)\n\n# [MCA IGNOU GROUP](https:\/\/mca.ignougroup.com\/)\nDownload Solved Assignments 2017-2018, Solved Question Papers Best Answers from Yahoo.com, Quora.com, Stackoverflow.com, Ignou.ac.in, Solved Assignments, Solution to any question\n\n- [Home](https:\/\/mca.ignougroup.com\/)\n- [Ask a Question](http:\/\/ask.ignougroup.com\/)\n- [Assignments](http:\/\/mca.ignougroup.com\/p\/assignments.html)\n  - [MCA 1st Sem](http:\/\/mca.ignougroup.com\/p\/assignments.html)\n  - [MCA 2nd Sem](http:\/\/mca.ignougroup.com\/p\/assignment-mca-2nd-semester-2015-2016.html)\n  - [MCA 3rd Sem](http:\/\/mca.ignougroup.com\/p\/ignou-mca-assignment-3rd-semester-2016.html)\n  - [MCA 4th Sem](http:\/\/mca.ignougroup.com\/p\/ignou-assignment.html)\n  - [MCA 5th Sem](http:\/\/mca.ignougroup.com\/p\/mca-5th-sem-solved-assignment.html)\n- [Question Papers](http:\/\/mca.ignougroup.com\/p\/question-paper.html)\n- [Contact](https:\/\/mca.ignougroup.com\/2017\/01\/why-is-it-best-to-use-prime-number-as.html)\n\n# Why is it best to use a prime number as a mod in a hashing function?\nposted by  ,\n\non\n\n7:42:00 AM\n\n,  [No Comments](https:\/\/mca.ignougroup.com\/2017\/01\/why-is-it-best-to-use-prime-number-as.html#comment-form)\n\n**Question Detail:**\n\nIf I have a list of key values from 1 to 100 and I want to organize them in an array of 11 buckets, I've been taught to form a mod function\n\n\\$\\$ H = k \\\\bmod \\\\ 11\\$\\$\n\nNow all the values will be placed one after another in 9 rows. For example, in the first bucket there will be \\$0, 11, 22 \\\\dots\\$. In the second, there will be \\$1, 12, 23 \\\\dots\\$ etc.\n\nLet's say I decided to be a bad boy and use a non-prime as my hashing function - take 12. Using the Hashing function\n\n\\$\\$ H = k \\\\bmod \\\\ 12\\$\\$\n\nwould result in a hash table with values \\$0, 12, 24 \\\\dots \\$ in the first bucket, \\$1, 13, 25 \\\\dots\\$ etc. in the second and so on.\n\nEssentially they are the same thing. I didn't reduce collisions and I didn't spread things out any better by using the prime number hash code and I can't see how it is ever beneficial.\n###### Asked By : [CodyBugstein](http:\/\/cs.stackexchange.com\/users\/4348)\n\n###### Best Answer from StackOverflow\nQuestion Source : <http:\/\/cs.stackexchange.com\/questions\/11029>\n#### Answered By : [Mario Cervera](http:\/\/cs.stackexchange.com\/users\/57681\/)\nConsider the set of keys \\$K=\\\\{0,1,...,100\\\\}\\$ and a hash table where the number of buckets is \\$m=12\\$. Since \\$3\\$ is a factor of \\$12\\$, the keys that are multiples of \\$3\\$ will be hashed to buckets that are multiples of \\$3\\$:\n\n- Keys \\$\\\\{0,12,24,36,...\\\\}\\$ will be hashed to bucket \\$0\\$.\n- Keys \\$\\\\{3,15,27,39,...\\\\}\\$ will be hashed to bucket \\$3\\$.\n- Keys \\$\\\\{6,18,30,42,...\\\\}\\$ will be hashed to bucket \\$6\\$.\n- Keys \\$\\\\{9,21,33,45,...\\\\}\\$ will be hashed to bucket \\$9\\$.\n\nIf \\$K\\$ is uniformly distributed (i.e., every key in \\$K\\$ is equally likely to occur), then the choice of \\$m\\$ is not so critical. But, what happens if \\$K\\$ is not uniformly distributed? Imagine that the keys that are most likely to occur are the multiples of \\$3\\$. In this case, all of the buckets that are not multiples of \\$3\\$ will be empty with high probability (which is really bad in terms of hash table performance).\n\nThis situation is more common that it may seem. Imagine, for instance, that you are keeping track of objects based on where they are stored in memory. If your computer's word size is four bytes, then you will be hashing keys that are multiples of \\$4\\$. Needless to say that choosing \\$m\\$ to be a multiple of \\$4\\$ would be a terrible choice: you would have \\$3m\/4\\$ buckets completely empty, and all of your keys colliding in the remaining \\$m\/4\\$ buckets.\n\nIn general:\n\n> Every key in \\$K\\$ that shares a common factor with the number of buckets \\$m\\$ will be hashed to a bucket that is a multiple of this factor.\n\nTherefore, to minimize collisions, it is important to reduce the number of common factors between \\$m\\$ and the elements of \\$K\\$. How can this be achieved? By choosing \\$m\\$ to be a number that has very few factors: a **prime number**.\n\n- — Share It —\n- [Facebook](http:\/\/www.facebook.com\/sharer.php?u=&title=Why%20is%20it%20best%20to%20use%20a%20prime%20number%20as%20a%20mod%20in%20a%20hashing%20function?)\n- [Twitter](http:\/\/twitter.com\/share?url=https:\/\/mca.ignougroup.com\/2017\/01\/why-is-it-best-to-use-prime-number-as.html&title=Why%20is%20it%20best%20to%20use%20a%20prime%20number%20as%20a%20mod%20in%20a%20hashing%20function?)\n- [Google+](https:\/\/plus.google.com\/share?url=https:\/\/mca.ignougroup.com\/2017\/01\/why-is-it-best-to-use-prime-number-as.html&title=Why%20is%20it%20best%20to%20use%20a%20prime%20number%20as%20a%20mod%20in%20a%20hashing%20function?)\n- [Linkedin](http:\/\/www.linkedin.com\/shareArticle?mini=true&url=https:\/\/mca.ignougroup.com\/2017\/01\/why-is-it-best-to-use-prime-number-as.html&title=Why%20is%20it%20best%20to%20use%20a%20prime%20number%20as%20a%20mod%20in%20a%20hashing%20function?)\n- [Pin It](http:\/\/pinterest.com\/pin\/create\/button\/?url=https:\/\/mca.ignougroup.com\/2017\/01\/why-is-it-best-to-use-prime-number-as.html&media=&description=Question%20Detail:%20%C2%A0%20If%20I%20have%20a%20list%20of%20key%20values%20from%201%20to%20100%20and%20I%20want%20to%20organize%20them%20in%20an%20array%20of%2011%20buckets,%20I've%20been%20taught%20...)\n#### [Akshay Mishra](https:\/\/mca.ignougroup.com\/2017\/01\/why-is-it-best-to-use-prime-number-as.html)\n\n[Newer Post](https:\/\/mca.ignougroup.com\/2017\/01\/convert-cfg-to-pda.html \"Newer Post\")   [Older Post](https:\/\/mca.ignougroup.com\/2017\/01\/why-are-some-programming-languages.html \"Older Post\")  [Home](https:\/\/mca.ignougroup.com\/)\n\n#### 0 comments:\n#### Post a Comment\nLet us know your responses and feedback\n\n## Search By Course Code\n## Blog Archive\n- [►]()  [2022](https:\/\/mca.ignougroup.com\/2022\/) (166)\n  - 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For example, in the first bucket there will be \\$0, 11, 22 \\\\dots\\$. In the second, there will be \\$1, 12, 23 \\\\dots\\$ etc.\n\nLet's say I decided to be a bad boy and use a non-prime as my hashing function - take 12. Using the Hashing function\n\n\\$\\$ H = k \\\\bmod \\\\ 12\\$\\$\n\nwould result in a hash table with values \\$0, 12, 24 \\\\dots \\$ in the first bucket, \\$1, 13, 25 \\\\dots\\$ etc. in the second and so on.\n\nEssentially they are the same thing. I didn't reduce collisions and I didn't spread things out any better by using the prime number hash code and I can't see how it is ever beneficial.\n###### Asked By : [CodyBugstein](http:\/\/cs.stackexchange.com\/users\/4348)\n#### Answered By : [Mario Cervera](http:\/\/cs.stackexchange.com\/users\/57681\/)\nConsider the set of keys \\$K=\\\\{0,1,...,100\\\\}\\$ and a hash table where the number of buckets is \\$m=12\\$. Since \\$3\\$ is a factor of \\$12\\$, the keys that are multiples of \\$3\\$ will be hashed to buckets that are multiples of \\$3\\$:\n\n- Keys \\$\\\\{0,12,24,36,...\\\\}\\$ will be hashed to bucket \\$0\\$.\n- Keys \\$\\\\{3,15,27,39,...\\\\}\\$ will be hashed to bucket \\$3\\$.\n- Keys \\$\\\\{6,18,30,42,...\\\\}\\$ will be hashed to bucket \\$6\\$.\n- Keys \\$\\\\{9,21,33,45,...\\\\}\\$ will be hashed to bucket \\$9\\$.\n\nIf \\$K\\$ is uniformly distributed (i.e., every key in \\$K\\$ is equally likely to occur), then the choice of \\$m\\$ is not so critical. But, what happens if \\$K\\$ is not uniformly distributed? Imagine that the keys that are most likely to occur are the multiples of \\$3\\$. In this case, all of the buckets that are not multiples of \\$3\\$ will be empty with high probability (which is really bad in terms of hash table performance).\n\nThis situation is more common that it may seem. Imagine, for instance, that you are keeping track of objects based on where they are stored in memory. If your computer's word size is four bytes, then you will be hashing keys that are multiples of \\$4\\$. Needless to say that choosing \\$m\\$ to be a multiple of \\$4\\$ would be a terrible choice: you would have \\$3m\/4\\$ buckets completely empty, and all of your keys colliding in the remaining \\$m\/4\\$ buckets.\n\nIn general:\n\n> Every key in \\$K\\$ that shares a common factor with the number of buckets \\$m\\$ will be hashed to a bucket that is a multiple of this factor.\n\nTherefore, to minimize collisions, it is important to reduce the number of common factors between \\$m\\$ and the elements of \\$K\\$. How can this be achieved? By choosing \\$m\\$ to be a number that has very few factors: a **prime number**.","meta_canonical":null}

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Boilerpipe Text	Question Detail: If I have a list of key values from 1 to 100 and I want to organize them in an array of 11 buckets, I've been taught to form a mod function $$ H = k \bmod \ 11$$ Now all the values will be placed one after another in 9 rows. For example, in the first bucket there will be $0, 11, 22 \dots$. In the second, there will be $1, 12, 23 \dots$ etc. Let's say I decided to be a bad boy and use a non-prime as my hashing function - take 12. Using the Hashing function $$ H = k \bmod \ 12$$ would result in a hash table with values $0, 12, 24 \dots $ in the first bucket, $1, 13, 25 \dots$ etc. in the second and so on. Essentially they are the same thing. I didn't reduce collisions and I didn't spread things out any better by using the prime number hash code and I can't see how it is ever beneficial. Asked By : CodyBugstein Answered By : Mario Cervera Consider the set of keys $K=\{0,1,...,100\}$ and a hash table where the number of buckets is $m=12$. Since $3$ is a factor of $12$, the keys that are multiples of $3$ will be hashed to buckets that are multiples of $3$: Keys $\{0,12,24,36,...\}$ will be hashed to bucket $0$. Keys $\{3,15,27,39,...\}$ will be hashed to bucket $3$. Keys $\{6,18,30,42,...\}$ will be hashed to bucket $6$. Keys $\{9,21,33,45,...\}$ will be hashed to bucket $9$. If $K$ is uniformly distributed (i.e., every key in $K$ is equally likely to occur), then the choice of $m$ is not so critical. But, what happens if $K$ is not uniformly distributed? Imagine that the keys that are most likely to occur are the multiples of $3$. In this case, all of the buckets that are not multiples of $3$ will be empty with high probability (which is really bad in terms of hash table performance). This situation is more common that it may seem. Imagine, for instance, that you are keeping track of objects based on where they are stored in memory. If your computer's word size is four bytes, then you will be hashing keys that are multiples of $4$. Needless to say that choosing $m$ to be a multiple of $4$ would be a terrible choice: you would have $3m/4$ buckets completely empty, and all of your keys colliding in the remaining $m/4$ buckets. In general: Every key in $K$ that shares a common factor with the number of buckets $m$ will be hashed to a bucket that is a multiple of this factor. Therefore, to minimize collisions, it is important to reduce the number of common factors between $m$ and the elements of $K$. How can this be achieved? By choosing $m$ to be a number that has very few factors: a prime number .
Markdown	[World's most popular travel blog for travel bloggers.](http://circularjourney.com/?utm_source=mca-blog) # [MCA IGNOU GROUP](https://mca.ignougroup.com/) Download Solved Assignments 2017-2018, Solved Question Papers Best Answers from Yahoo.com, Quora.com, Stackoverflow.com, Ignou.ac.in, Solved Assignments, Solution to any question - [Home](https://mca.ignougroup.com/) - [Ask a Question](http://ask.ignougroup.com/) - [Assignments](http://mca.ignougroup.com/p/assignments.html) - [MCA 1st Sem](http://mca.ignougroup.com/p/assignments.html) - [MCA 2nd Sem](http://mca.ignougroup.com/p/assignment-mca-2nd-semester-2015-2016.html) - [MCA 3rd Sem](http://mca.ignougroup.com/p/ignou-mca-assignment-3rd-semester-2016.html) - [MCA 4th Sem](http://mca.ignougroup.com/p/ignou-assignment.html) - [MCA 5th Sem](http://mca.ignougroup.com/p/mca-5th-sem-solved-assignment.html) - [Question Papers](http://mca.ignougroup.com/p/question-paper.html) - [Contact](https://mca.ignougroup.com/2017/01/why-is-it-best-to-use-prime-number-as.html) # Why is it best to use a prime number as a mod in a hashing function? posted by , on 7:42:00 AM , [No Comments](https://mca.ignougroup.com/2017/01/why-is-it-best-to-use-prime-number-as.html#comment-form) Question Detail: If I have a list of key values from 1 to 100 and I want to organize them in an array of 11 buckets, I've been taught to form a mod function \$\$ H = k \\bmod \\ 11\$\$ Now all the values will be placed one after another in 9 rows. For example, in the first bucket there will be \$0, 11, 22 \\dots\$. In the second, there will be \$1, 12, 23 \\dots\$ etc. Let's say I decided to be a bad boy and use a non-prime as my hashing function - take 12. Using the Hashing function \$\$ H = k \\bmod \\ 12\$\$ would result in a hash table with values \$0, 12, 24 \\dots \$ in the first bucket, \$1, 13, 25 \\dots\$ etc. in the second and so on. Essentially they are the same thing. I didn't reduce collisions and I didn't spread things out any better by using the prime number hash code and I can't see how it is ever beneficial. ###### Asked By : [CodyBugstein](http://cs.stackexchange.com/users/4348) ###### Best Answer from StackOverflow Question Source : <http://cs.stackexchange.com/questions/11029> #### Answered By : [Mario Cervera](http://cs.stackexchange.com/users/57681/) Consider the set of keys \$K=\\{0,1,...,100\\}\$ and a hash table where the number of buckets is \$m=12\$. Since \$3\$ is a factor of \$12\$, the keys that are multiples of \$3\$ will be hashed to buckets that are multiples of \$3\$: - Keys \$\\{0,12,24,36,...\\}\$ will be hashed to bucket \$0\$. - Keys \$\\{3,15,27,39,...\\}\$ will be hashed to bucket \$3\$. - Keys \$\\{6,18,30,42,...\\}\$ will be hashed to bucket \$6\$. - Keys \$\\{9,21,33,45,...\\}\$ will be hashed to bucket \$9\$. If \$K\$ is uniformly distributed (i.e., every key in \$K\$ is equally likely to occur), then the choice of \$m\$ is not so critical. But, what happens if \$K\$ is not uniformly distributed? Imagine that the keys that are most likely to occur are the multiples of \$3\$. In this case, all of the buckets that are not multiples of \$3\$ will be empty with high probability (which is really bad in terms of hash table performance). This situation is more common that it may seem. Imagine, for instance, that you are keeping track of objects based on where they are stored in memory. If your computer's word size is four bytes, then you will be hashing keys that are multiples of \$4\$. Needless to say that choosing \$m\$ to be a multiple of \$4\$ would be a terrible choice: you would have \$3m/4\$ buckets completely empty, and all of your keys colliding in the remaining \$m/4\$ buckets. In general: > Every key in \$K\$ that shares a common factor with the number of buckets \$m\$ will be hashed to a bucket that is a multiple of this factor. Therefore, to minimize collisions, it is important to reduce the number of common factors between \$m\$ and the elements of \$K\$. How can this be achieved? 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Readable Markdown	Question Detail: If I have a list of key values from 1 to 100 and I want to organize them in an array of 11 buckets, I've been taught to form a mod function \$\$ H = k \\bmod \\ 11\$\$ Now all the values will be placed one after another in 9 rows. For example, in the first bucket there will be \$0, 11, 22 \\dots\$. In the second, there will be \$1, 12, 23 \\dots\$ etc. Let's say I decided to be a bad boy and use a non-prime as my hashing function - take 12. Using the Hashing function \$\$ H = k \\bmod \\ 12\$\$ would result in a hash table with values \$0, 12, 24 \\dots \$ in the first bucket, \$1, 13, 25 \\dots\$ etc. in the second and so on. Essentially they are the same thing. I didn't reduce collisions and I didn't spread things out any better by using the prime number hash code and I can't see how it is ever beneficial. ###### Asked By : [CodyBugstein](http://cs.stackexchange.com/users/4348) #### Answered By : [Mario Cervera](http://cs.stackexchange.com/users/57681/) Consider the set of keys \$K=\\{0,1,...,100\\}\$ and a hash table where the number of buckets is \$m=12\$. Since \$3\$ is a factor of \$12\$, the keys that are multiples of \$3\$ will be hashed to buckets that are multiples of \$3\$: - Keys \$\\{0,12,24,36,...\\}\$ will be hashed to bucket \$0\$. - Keys \$\\{3,15,27,39,...\\}\$ will be hashed to bucket \$3\$. - Keys \$\\{6,18,30,42,...\\}\$ will be hashed to bucket \$6\$. - Keys \$\\{9,21,33,45,...\\}\$ will be hashed to bucket \$9\$. If \$K\$ is uniformly distributed (i.e., every key in \$K\$ is equally likely to occur), then the choice of \$m\$ is not so critical. But, what happens if \$K\$ is not uniformly distributed? Imagine that the keys that are most likely to occur are the multiples of \$3\$. In this case, all of the buckets that are not multiples of \$3\$ will be empty with high probability (which is really bad in terms of hash table performance). This situation is more common that it may seem. Imagine, for instance, that you are keeping track of objects based on where they are stored in memory. If your computer's word size is four bytes, then you will be hashing keys that are multiples of \$4\$. Needless to say that choosing \$m\$ to be a multiple of \$4\$ would be a terrible choice: you would have \$3m/4\$ buckets completely empty, and all of your keys colliding in the remaining \$m/4\$ buckets. In general: > Every key in \$K\$ that shares a common factor with the number of buckets \$m\$ will be hashed to a bucket that is a multiple of this factor. Therefore, to minimize collisions, it is important to reduce the number of common factors between \$m\$ and the elements of \$K\$. How can this be achieved? By choosing \$m\$ to be a number that has very few factors: a prime number.
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