{"id":5286,"date":"2022-04-08T00:01:13","date_gmt":"2022-04-08T07:01:13","guid":{"rendered":"https:\/\/c-for-dummies.com\/blog\/?p=5286"},"modified":"2022-04-02T11:44:57","modified_gmt":"2022-04-02T18:44:57","slug":"the-babylonian-square-root-solution","status":"publish","type":"post","link":"https:\/\/c-for-dummies.com\/blog\/?p=5286","title":{"rendered":"The Babylonian Square Root – Solution"},"content":{"rendered":"

\"Hammurabi\"
\nI was blown away by the elegance and simplicity of the Babylonian Method to calculate a square root, which is the topic of this month’s Exercise<\/a>. I only hope that my solution lives up the challenge, lest I suffer the fate of the Assyrians.
\n
\nThe challenge is to write the babylonian_sr()<\/em> function, which consumes a double<\/em> value and returns a double<\/em> value, a square root. I set a precision value of seven and use the passed argument as the “high” value, 1.0 as the initial “low” value; refer to the original post if you’re confused by these labels. Here is my solution:<\/p>\n

2022_04-Exercise.c<\/a><\/h3>\n
\r\n#include <stdio.h>\r\n\r\ndouble babylonian_sr(double root)\r\n{\r\n    double low,high;\r\n    int x;\r\n    const int precision = 7;\r\n\r\n    low = 1.0;\r\n    high = root;\r\n    for( x=0; x<precision; x++ )\r\n    {\r\n        high = (high+low)\/2.0;\r\n        low = root\/high;\r\n    }\r\n    return(low);\r\n}\r\n\r\nint main()\r\n{\r\n    double pv,sr;\r\n\r\n    printf(\"Enter a positive value: \");\r\n    scanf(\"%lf\",&pv);\r\n    if( pv <= 0 )\r\n        return(1);\r\n    sr = babylonian_sr(pv);\r\n    printf(\"The square root of %.0f is %f\\n\",\r\n            pv,\r\n            sr\r\n          );\r\n\r\n    return(0);\r\n}<\/pre>\n
My babylonian_sr()<\/em> function uses variable names stolen from Figure 1 in the Exercise post: root<\/code> is the value passed; low<\/code> and high<\/code> are the less-than and greater-than values used in the average calculation.<\/p>\n
The low<\/code> value is set to 1.0 at Line 9.<\/p>\n
The high<\/code> value is set equal to root<\/code> at Line 10. Remember, root<\/code> doesn’t change throughout the function. (It could be declared const<\/em>, but that’s just more typing.)<\/p>\n
Variable x<\/code> in the for<\/em> loop sets the precision, the number of times the two steps repeat. I chose the precision value 7, set at Line 7.<\/p>\n
Step 1 takes place at Line 13. The new high<\/code> value is determined as the average of high<\/code> and low<\/code>.<\/p>\n
Step 2 takes place at Line 14. The new low<\/code> value is calculated as the original number, root<\/code>, divided by high<\/code>.<\/p>\n
The process repeats precision<\/code> times, and the result stored in low<\/code> is returned at Line 16.<\/p>\n
Here’s a sample run:<\/p>\n
Enter a positive value: 5
\nThe square root of 5 is 2.236068<\/code><\/p>\n
For some large values, the result may not be perfect. If you find such an issue, increase the value of variable precision<\/code> to something greater than seven, but keep in mind that floating point values aren’t precise.<\/p>\n
If you really want to test your babylonian_sr()<\/em> function solution, add statements in the main()<\/em> function that compare your function’s output with C’s sqrt()<\/em> function. Remember to include the math.h<\/code> header, and if you use Linux, you must link in the m<\/code> library to build the code.<\/p>\n","protected":false},"excerpt":{"rendered":"
I was blown away by the elegance and simplicity of the Babylonian Method to calculate a square root, which is the topic of this month’s Exercise. I only hope that my solution lives up the challenge, lest I suffer the … Continue reading →<\/span><\/a><\/p>\n","protected":false},"author":1,"featured_media":0,"comment_status":"open","ping_status":"closed","sticky":false,"template":"","format":"standard","meta":{"footnotes":""},"categories":[5],"tags":[],"class_list":["post-5286","post","type-post","status-publish","format-standard","hentry","category-solution"],"_links":{"self":[{"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/5286","targetHints":{"allow":["GET"]}}],"collection":[{"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts"}],"about":[{"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/types\/post"}],"author":[{"embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/users\/1"}],"replies":[{"embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcomments&post=5286"}],"version-history":[{"count":3,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/5286\/revisions"}],"predecessor-version":[{"id":5303,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/5286\/revisions\/5303"}],"wp:attachment":[{"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=5286"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=5286"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=5286"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}