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Fredrik Robertsen
2025-08-13 10:49:47 +02:00
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# part 1
[solution](https://uiua.org/pad?src=0_14_0-dev_5__JmZyYXMgImlucHV0IgoKIyB0aGUgcmVnZXggZXhwcmVzc2lvbiBzdHJpbmcKIyB0byBtYXRjaCB0aGUgbXVsIGluc3RydWN0aW9ucwojIGFuZCBjYXB0dXJlIHRoZSBvcGVyYW5kcwpNdWwg4oaQICQgbXVsXCgoXGQrKSwoXGQrKVwpCgojIG1hdGNoIHRoZSBtdWwgZXhwcmVzc2lvbiwKIyBkcm9wIHRoZSBmdWxsICJtdWwoLi4uKSIgc3RyaW5nLCAKIyBwYXJzZSBlYWNoIG9wZXJhbmQgdG8gaW50ClBhcnNlIOKGkCDiiLXii5Ug4omh4oaYMSByZWdleCBNdWwKCiMgbXVsdGlwbHkgdGhlIG51bWJlcnMgaW4gZWFjaCByb3csCiMgc3VtIHRoZSBwcm9kdWN0cwpTb2wg4oaQIC8rIOKJoS_DlwoKU29sIFBhcnNlCg==)
i first solved it this morning using vim, since i had no idea how to use regex in uiua. this is how i did it in vim (with some regex help from [kagi assistant](https://help.kagi.com/kagi/ai/assistant.html)):
```
:e "input"
100J # there are few lines, join them all
:%s/\(mul(\d\+,\d\+)\)\|\(.\)/\1/g # find all mul expressions
:%s/mul(\(\d+\),\(\d+\))/\1*\2+/g # replace all mul expressions with the operands multiplied together with a + appended
$x # remove last +
:'<,'>s/.*/\=eval(submatch(0)) # evaluate the expression
```
my uiua solution is very similar