2 files changed, 0 insertions, 105 deletions
diff --git a/sem1/algo/mm3/multiply.c b/sem1/algo/mm3/multiply.c
deleted file mode 100644
index e454de3..0000000
--- a/sem1/algo/mm3/multiply.c
+++ /dev/null
@@ -1,56 +0,0 @@
-#include <stdio.h>
-#include <stdlib.h>
-#include <stdint.h>
-#include <time.h>
-
-uint64_t mult(uint64_t x, uint64_t y, unsigned int bits) {
-	/* Check if they expect more than 64 bits. */
-	if( bits > 64 ) {
-		printf("Sorry we cant handle higher than 64 bits\n");
-		exit(1);
-	} else if(bits <= 1) {
-		return x && y;
-	}
-
-	unsigned int newBits = bits >> 1;
-
-#if DO_PRINT == 1
-	printf("\n\n bits: %u, New bits: %u\n", bits, newBits);
-#endif 
-
-	/* Split up into to */
-	uint32_t halvMask = ((uint64_t)1 << newBits) - 1;
-#if DO_PRINT == 1
-	printf("Using mask 0x%08X\n", halvMask);
-#endif 
-	uint32_t a = (x >> (newBits)) & halvMask;
-	uint32_t b = x & halvMask;
-	uint32_t c = (y >> (newBits)) & halvMask;
-	uint32_t d = y & halvMask;
-
-#if DO_PRINT == 1
-	printf("A: 0x%08X, B: 0x%08X, C: 0x%08X, D: 0x%08X\n", a, b, c, d);
-#endif 
-
-	return (mult(a, c, newBits) << bits) + ((mult(a, d, newBits) + mult(b, c, newBits)) << newBits) + mult(b, d, newBits);
-
-}
-
-
-int main(void) {
-
-	uint32_t a = 0xFFFFFF;
-	uint8_t  b = 55;
-	uint64_t res;
-
-	clock_t begin = clock();
-	res = mult(a, b, 64);
-	clock_t end = clock();
-
-	printf("Result: %lld\n", res);
-
-	clock_t diff = end - begin;
-	printf("Took %d which is %f s\n", diff, (double)diff/CLOCKS_PER_SEC);
-	
-	return 0;
-}
diff --git a/sem1/algo/mm3/opgaver.md b/sem1/algo/mm3/opgaver.md
deleted file mode 100644
index a6e560c..0000000
--- a/sem1/algo/mm3/opgaver.md
+++ /dev/null
@@ -1,49 +0,0 @@
-# Opgave 1
-
-## Exercise 9.2
-
-**T(n) = 3T(n/2) + n**
-
-a = 3
-b = 2
-d(n) = n
-
-a ? d(b) -> 3 > 2
-
-Svaret er derfor O(n^log2(3)) = O(n^1.59)
-
-**T(n) = 3T(n/2) + n^2**
-
-3 ? 2^2 -> 3 < 4
-
-d(n) = n^2 hvilket betyder at O(n^2)
-
-**T(n) = 8T(n/2) + n^2**
-
-8 ? 2^3 -> 8 = 8
-
-d(n) = n^3 hvilket betyder at O(n^3 log2(n))
-
-## Evercise 9.3
-
-**T(n) = 4T(n/3) + n**
-
-a = 4
-b = 3
-d(n) = n
-
-4 ? 3 -> 4 > 3
-
-Derfor er svaret O(n^log3(4)) = O(n^1.26)
-
-**T(n) = 4T(n/3) + n^2**
-
-4 ? 3^2 -> 4 < 9
-
-d(n) = n^2 hvilket betyder at O(n^2)
-
-**T(n) = 9T(n/3) + n^2**
-
-9 ? 3^2 -> 9 = 9
-
-d(n) = n^2 hvilket betyder at O(n^2 log2(n))