{"id":6650,"date":"2024-11-08T00:01:07","date_gmt":"2024-11-08T08:01:07","guid":{"rendered":"https:\/\/c-for-dummies.com\/blog\/?p=6650"},"modified":"2024-11-02T11:39:02","modified_gmt":"2024-11-02T18:39:02","slug":"ethiopian-multiplication-solution","status":"publish","type":"post","link":"https:\/\/c-for-dummies.com\/blog\/?p=6650","title":{"rendered":"Ethiopian Multiplication – Solution"},"content":{"rendered":"

The challenge for this month’s Exercise<\/a> is to write code that uses the Ethiopian Multiplication method. The process involves doubling and halving the factors, then eliminating and finally tallying the result.
\n
\nFor my solution I use only a single while<\/em> loop. This loop both tallies the results and doubles\/halves the values. As I worked on the code, I discovered that you don’t really need to maintain a two-column list, which would involve more storage, but you can eliminate even value results in the second column as they’re calculated.<\/p>\n

Here is my solution:<\/p>\n

2024_11-Exercise.c<\/a><\/h3>\n
\r\n#include <stdio.h>\r\n\r\nint main()\r\n{\r\n    int a,b,total;\r\n\r\n    \/* gather input *\/<\/span>\r\n    printf(\"First value: \");\r\n    scanf(\"%d\",&a);\r\n    printf(\"Second value: \");\r\n    scanf(\"%d\",&b);\r\n\r\n    \/* do math *\/<\/span>\r\n    printf(\"Traditional: %d * %d = %d\\n\",a,b,a*b);\r\n    printf(\"  Ethiopian: %d * %d = \",a,b);\r\n    total = 0;\r\n    while(a>0)\r\n    {\r\n        total += a%2 ? b : 0;\r\n        a>>=1;\r\n        b<<=1;\r\n    }\r\n    printf(\"%d\\n\",total);\r\n\r\n    return 0;\r\n}<\/pre>\n
Variable a<\/code> is the first value, the first column. Variable b<\/code> is the second. I’m unsure whether a specific order is necessary. I didn’t test Ethiopian Multiplication to see if swapping the numbers changes anything. (I don’t think it would.)<\/p>\n
The while<\/em> loop spins until the value of a<\/code> is zero or less. Variable total<\/code> calculates the result, accumulating values as they’re manipulated.<\/p>\n
The ternary operator tests to see whether variable a<\/code> (the first column) is even:<\/p>\n
a%2 ? b : 0<\/code><\/p>\n
If so, zero is added to variable total<\/code>. Otherwise, the value of variable b<\/code> is added to total<\/code>.<\/p>\n
Next, variable a<\/code> is halved: a>>=1<\/code><\/p>\n
Then variable b<\/code> is doubled: b<<=1<\/code><\/p>\n
When the loop terminates, the result is output. It’s matched with the result generated from using the *<\/code> operator.<\/p>\n
Here’s a sample run:<\/p>\n
First value: 22
\nSecond value: 47
\nTraditional: 22 * 47 = 1034
\n  Ethiopian: 22 * 47 = 1034<\/code><\/p>\n
I hope that your solution met with success! Remember, it need not be identical to mine. In fact, I can think of several ways to perform this calculation. As long as it uses the Ethiopian methodology and calculates the same result, consider your solution valid and passing the challenge.<\/p>\n","protected":false},"excerpt":{"rendered":"
The challenge for this month’s Exercise is to write code that uses the Ethiopian Multiplication method. The process involves doubling and halving the factors, then eliminating and finally tallying the result.<\/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-6650","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\/6650","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=6650"}],"version-history":[{"count":5,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/6650\/revisions"}],"predecessor-version":[{"id":6672,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=\/wp\/v2\/posts\/6650\/revisions\/6672"}],"wp:attachment":[{"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fmedia&parent=6650"}],"wp:term":[{"taxonomy":"category","embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Fcategories&post=6650"},{"taxonomy":"post_tag","embeddable":true,"href":"https:\/\/c-for-dummies.com\/blog\/index.php?rest_route=%2Fwp%2Fv2%2Ftags&post=6650"}],"curies":[{"name":"wp","href":"https:\/\/api.w.org\/{rel}","templated":true}]}}