The fewest number of cuts will be 2.

If we cut it into 3 part where the parts are 1, 2 and 4 segments long.
Day 1 – Give 1 segment bar
Day 2 – Give 2 segment bar ask for 1 segment bar back
Day 3 – Give 1 segment bar
Day 4 – Give 4 segment bar ask for 1 + 2 back
Day 5 – Give 1 segment bar
Day 6 – Give 2 segment bar ask for 1 segment bar back
Day 7 – Give 1 segment bar

I am a digital hardware engineer — Here goes my solution

i’ll require 3 bit to give id to each bar 1st
000
001
010
011
100
101
110
111

Here if you know each bit represent weights of 4, 2 and 1 i.e, 111 means 4*1+2*1+1*1 = 7

Now you know min bar length of 4, 2  and 1 we can use to represent / distribute all 7 pieces.

421
001 –> give bar1 1st day (weight1 is 1 and rest are 0)
010 –> take bar1 back (see weight1 is 0) and give bar 2 (weight2 is 1 )
011 –> give bar1 back again
100 –> take both bar1 , bar2 and give bar 4.
101 –> give bar1 again.
110 –> take bar1 back and give bar2 again.
111 –> give bar1 as well. We are done 🙂

Just 2 cuts.
First one between 1 and 2 and second between 3 and 4.
lets number them from 1 to 7
So you have 1   2-3   4-5-6-7 in this formation and connection
Day 1- Give 1
Day 2- Give 2-3 take back 1
Day 3- Give 1 also
Day 4- Take all back and give 4-5-6-7
Day 5- Give 1 with the rest
Day 6- Take back 1 and give 2-3
Day 7- Give 1.

So we need only two cuts