So the scores you lost for bad/partial answer were double. ]]>

Note that it's not just that there is a partition iff there is a XS, but the groups themselves are the same (the group A from partition is exactly the sub-group S in XS, with the only difference of +1 to each member of S, and the group B is everything not in S).

]]>one side of the proof is trivial (perhaps trivial isn't the most appropriate definition)

but how can you prove that if there isn't a partition, there can't be a XS ?

]]>and it seems he was right…. ]]>

for {0,0,0,1} there is no partition

the reduction {0,0,0,1,1,1,1,2}

for your new group choose (repetitions ARE ALLOWED) {1,1,2,0,0} {1,1,0}

both sides of the equation will sum up to 7.

so one side of the "if" is correct but "iff" isn't.

]]>PARTITION<XS:

f({a1,…an})={x1,…,xn}

such that: xi = ai+1

תציבו את ה- xi

בסכום של XS

PARTITIONותראו שחוזרים לסכום של

כך שמתקיים אמ"מ כנדרש

1,4,7,0,0,0 belong to XS

1,4,

7,0,0,0

but XS({1,3,0,0})=TRUE

S={1}

1+ 3 = (3+0+0) +1

]]>PARTITION<XS:

f({X1,…Xn})={X1,…,Xn,0,0…0} (add n zeros).

]]>Some one can publish his reduction for question number 1? (if he think he's right)

Thanks :)

]]>